Application of DerivativeHard
Question
The set of values of p for which the extremum of the function f(x) = x3 - 3 px2 + 3 (p2 - 1) x + 1 lie in the interval (- 2, 4), is:
Options
A.(- 3, 5)
B.(- 3, 3)
C.(- 1, 3)
D.(- 1, 4)
Solution
f′(x) = 3x2 - 3p2x + 3p2 - 3
= 3((x - p)2 - 1)
= 3(x - (p + 1)) (x - (p - 1))
⇒ p - 1 > - 2 and p + 1 < 4
⇒ p > - 1 and p < 3
⇒ - 1 < p < 3
= 3((x - p)2 - 1)
= 3(x - (p + 1)) (x - (p - 1))
⇒ p - 1 > - 2 and p + 1 < 4
⇒ p > - 1 and p < 3
⇒ - 1 < p < 3
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
The first term of an infinitely decreasing G.P. is unity and its sum is S. The sum of the squares of the terms of the pr...The rate of change of the area of a circular disc with respect to its circumference when the radius is 3 cm, is-...The equation of tangent to the curve y = 2 sin x + sin 2x at the point x = π/3 is-...The distance between the origin and the normal to the curve y = e2x + x2 at the point x = 0 is-...The coordinates fo points(s) at each of which the tangents to the curve y = x3 - 3x2 - 7x + 6 cut off on the positive se...